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3 Responses
Great instinct to think about the graph going on forever-but “forever” can weave around trouble spots! For f(x) = 1/(x − 2), the only input that’s forbidden is x = 2, because the denominator would be zero; so the domain is all real numbers except 2 (not 0). You can see this as the classic 1/x graph shifted right by 2, which moves the vertical asymptote from x = 0 to x = 2. For the range, your idea is spot on: f(x) can never be 0, since 1/(x − 2) = 0 has no solution (a reciprocal is never zero), so the range is all real numbers except 0. Another neat check is to solve y = 1/(x − 2) for x, giving x = 2 + 1/y, which works for every y except y = 0-confirming the horizontal asymptote at y = 0. As a follow-up, what happens to the domain, range, and asymptotes if we look at g(x) = 1/(x − 2) + 3?
You’re really close-since 1/(x−2) is just 1/x slid sideways, the only input it can’t take is x=0, so the domain is all reals except 0. And because a reciprocal never hits 0, the range is all reals except 0 too-I’m pretty sure that’s right.
The only place this function misbehaves is where you’d be dividing by zero, i.e., when x−2=0. So the domain is all real numbers except x=2. x=0 is totally fine: f(0)=1/(0−2)=−1/2. For the range, notice 1/(x−2) can never be 0, because a nonzero numerator can’t produce 0-so the range is all real numbers except 0. Think vertical asymptote at x=2 (blows up there) and horizontal asymptote at y=0 (gets close but never hits). Quick example: if you want f(x)=5, solve 1/(x−2)=5, so x−2=1/5 and x=2.2. That’s fine. Try to make f(x)=0 and you’ll get 1/(x−2)=0, which has no solution. So: domain = all reals except 2; range = all reals except 0.